Is the closure of a set the same as its smallest closed set containing it?

[Bipolarity]

My first analysis/topology text defined the boundary of a set S as the set of all points whose neighborhoods had some point in the set S and some point outside the set S. It also defined the closure of a set S the union of S and its boundary.

Using this, we can prove that the closure of S is the smallest closed set containing S. We can also prove that the boundary of S is the intersection of the closure of S and the closure of the complement of S.

I was wondering, if we define the closure of S to be the smallest closed set containing S, and the boundary of S to be the intersection of the closure of S and the closure of the complement of S, will we have the machinery necessary to work backwards and prove the first two definitions, i.e. are the two definitions of boundary/closure equivalent?

BiP

 

[economicsnerd]

Yes, they are! And your proof will show it!

Given a subset S of your space X...
- Let \partial S=\{x\in X: \enspace \text{ for any neighborhood } N \text{ of } x, \enspace N\cap S\neq\emptyset \text{ and } N\cap (X\setminus S)\neq\emptyset \}.
- Let \bar S = S \cup \partial S.
- Let \tilde S be the smallest closed set containing S. (You have to prove that there is such a set. But this is easy; it's just the intersection of every closed set containing S.)

Notice that I haven't used the words "closure" or "boundary" anywhere above.

It sounds like you know how to prove that \bar S = \tilde S and \partial S = \bar S \cap \overline{X\setminus S}.

Having shown that, you can define:
- The closure of S (denoted cl(S)) is either \bar S or \tilde S, whichever definition you like. [We now know they're equivalent.]
- The boundary of S is either \partial S or cl(S)\cap cl(X\setminus S), whichever definition you like. [We now know they're equivalent.]